Puzzles
16 December
Arrange the digits 1-9 in a 3×3 square so that the first row makes a triangle number, the second row's digits are all even, the third row's digits are all odd; the first column makes a square number, and the second column makes a cube number.
The number in the third column is today's number.
triangle | |||
all digits even | |||
all digits odd | |||
square | cube | today's number |
15 December
Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.
14 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10. Today's number is the product of the numbers in the red boxes.
- | + | = 10 | |||
÷ | + | ÷ | |||
÷ | + | = 3 | |||
+ | - | ÷ | |||
+ | × | = 33 | |||
= 7 | = 3 | = 3 |
13 December
There is a row of 1000 lockers numbered from 1 to 1000. Locker 1 is closed and locked and the rest are open.
A queue of people each do the following (until all the lockers are closed):
- Close and lock the lowest numbered locker with an open door.
- Walk along the rest of the queue of lockers and change the state (open them if they're closed and close them if they're open) of all the lockers that are multiples of the locker they locked.
Today's number is the number of lockers that are locked at the end of the process.
Note: closed and locked are different states.
12 December
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometimes the three vertices you pick form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.
11 December
This puzzle is inspired by a puzzle Woody showed me at MathsJam.
Today's number is the number \(n\) such that $$\frac{216!\times215!\times214!\times...\times1!}{n!}$$ is a square number.
10 December
The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.
9 December
Today's number is the number of numbers between 10 and 1,000 that contain no 0, 1, 2 or 3.